Consider the following circuit. Due to reverse battery installation, accidental short circuiting in the supply side, or other inappropriate operation will damage the load/Equipment. By inserting a normal PN junction diode or schottky diode in series with battery will provide the reverse battery protection.
But this method will resulting in relatively higher power loss(in case of normal PN junction diode) or more expensive( in case of schottky diode method).

Method # 3 - Reverse Voltage Protection using MOSFET

In place of the diode, we can use a P-channel MOSFET with the drain terminal connected to the battery, the source terminal connected to the load, and gate terminal connected to the ground.

As shown in the figure, the MOSFET has a in-built body diode.

Initially it will behave just like the series diode.

So the current will flow, momentarily the voltage will drop across the diode( ie, forward voltage drop ([VF]).

If the voltage at the source terminal exceeds the Gate-Source threshold voltage(VGS[Th]) of the MOSFET, then the MOSFET will turn on.

So the current will now flow from drain to source without any voltage drop.

If the battery is connected in reverse manner, the in-built body diode is reverse biased and the gate voltage exceeds the source voltage.

So no current will flow.

Case Study:

For example, assume a P-channel MOSFET with a forward voltage drop(VF) of 0.7 volts, RDS = 0.1 Ohms and a Gate-Source threshold voltage(VGS[Th]) of 1.5 volts is used for the reverse voltage protection.
ie,
VF = 0.7 Volts
VGS[Th] = 1.5 Volts
RDS = 0.1 Ohms
If you connect a 3.3 volt supply, the load will initially receive (3.3 - 0.7) = 2.6 volts.
At this moment,
VSOURCE = 2.6V
VGATE = 0 Volts
Threshold VGS = 1.5 V
Actual VGS = 2.6V - 0V = 2.6V
As the actual VGS exceeds the threshold VGS, the MOSFET will turn on almost instantly. Consequently shorts out the body diode and now provides almost the entire 3.3 volts to the load.
After that the actual power loss will be due to the RDS of the MOSFET. If the load draws 200mA current then the power loss will be
Power loss = I2R
= (0.2 x 0.2 x 0.1)
= 0.004W
= 4mW
Which is the lowest power loss compare to junction diode, schottky diode based protection...

Method # 3A - Reverse Voltage Protection using Low VDS MOSFET

In the previous method, the Mosfet should be chosen such that the VDS value will be more than the Supply voltage.

In many cases, the voltage level will be in the range of few hundreds. MOSFETs having high VDS is more expensive.

To overcome this issue, a zener diode and a series resistor have to be connected as shown in the figure.

The zener will limit the actual VGS well below the threshold VGS.

The zener diode should be selected such that its zener voltage is less than threshold VGS.